 | Robert Simson - 1806 - 546 Seiten
...in which the vertex of one triangle is upon a side of the other, needs no demonstration. Therefore upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise... | |
 | John Playfair - 1806 - 320 Seiten
...in which the vertex of one triangle is upon a side of the other, needs no demonstration. Therefore, upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise... | |
 | John Mason Good - 1813 - 714 Seiten
...which subtend, or arc. opposite to» the equal angles, shall be equal to one another. Prop. VII. Theor. Upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise... | |
 | Charles Butler - 1814 - 528 Seiten
..." for if -4EB do not coincide with CFD, it must fall otherwise (as in the figure to prop. 23.) then upon the same base, and on the same side of it, there will be two similar segments of circles not coinciding with one another, but this has been shewn (in... | |
 | Euclides - 1816 - 588 Seiten
...in which the vertex of one triangle is upon a side of the other, needs no demonstration. Therefore, upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise... | |
 | John Playfair - 1819 - 354 Seiten
...two angles, &c. QED II C COR. Hence every equiangular triangle is also equilateral. PROP. VII. THEOR. Upon the. same base, and on the same side of it, there caitnot be two triangles, that have their sides which are terminated in one extremity of the base equal... | |
 | Euclides - 1821 - 294 Seiten
...every equiangular triangle is equilateral ; vide, Elrington. PROP. 7. THEOR. i On the same right line and on the same side of it there cannot be two triangles formed whose conterminous sides are equal. If it be possible that there can, 1st, let the vertex of... | |
 | Rev. John Allen - 1822 - 508 Seiten
...it are equal, and therefore the sides opposite to them. PROP. VII. THEOR. Upon the same base (AB), and on the same side of it, there cannot be two triangles (ACB, ADB), whose conterminous sides are equal, (namely AC to AD, and BC to BD). For, if possible,... | |
 | Peter Nicholson - 1825 - 1046 Seiten
...&c. QED COR. Hence every equiangular triangle is also equal equilátera. Proposition Vll. Theorem. Upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another ; and... | |
 | Robert Simson - 1827 - 546 Seiten
...two angles, &c. QED COR. — Hence every equiangular triangle is also equilateral. PROP. VII. THEOR. Upon the same base, and on the same side of it, there can- See N. not be two triangles that have their sides which are terminated in one extremity of the... | |
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Upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of' the base, equal to one another, and likewise those which are terminated in the other extremity. 
